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denis-greek [22]
3 years ago
14

Round 824.0472 to the nearest hundredth.

Mathematics
1 answer:
Gelneren [198K]3 years ago
4 0

Answer:

824.05

Step-by-step explanation:

the hundreths place is two spots behind the decimal, there is a four. Look at the number next to it is a 7 so you round up

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The following series are geometric series or a sum of two geometric series. Determine whether each series converges or not. For
Sergeeva-Olga [200]

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Required solution gives series (a) divergent, (b) convergent, (c) divergent.

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(a) Given,

\sum_{n\to 0}^{\infty}\frac{2^n}{9^{2n}+1}

To applying limit comparison test, let  a_n=\frac{2^n}{9^{2n}+1} and b_n=\frac{9^{2n}}{2^n}. Then,

\lim_{n\to\infty} \frac{a_n}{b_n}=\lim_{n\to\infty}(1+\frac{1}{9^{2n}})=1>0

Because of the existance of limit and the series  \frac{9^{2n}}{2^n} is divergent since \frac{9^{2n}}{2^n}=(\frac{9^2}{2})^n where \frac{81}{2}>1, given series is divergent.  

(b) Given,

\sum_{n\to 1}^{\infty}(\frac{7^n}{7^n+4})

Again to apply limit comparison test let a_n=\frac{7^n}{7^n+4} and b_n=\frac{1}{7^n} we get,

\lim_{n\to \infty}\frac{a_n}{b_n}=\frac{1}{7^n+4}=0

Since \lim_{n\to \infty} \frac{1}{7^n}=0 is convergent, by comparison test, given series is convergent.

(c) Given,

\sum_{n\to 1}^{\infty}\frac{5^n+2^n}{6^n}= \sum_{n\to 1}^{\infty}(\frac{5}{6})^n+\sum_{n\to 1}^{\infty}(\frac{1}{3})^n . Now applying Cauchy Root test on last two series, we will get,

  • \lim_{n\to \infty}|(\frac{5}{6})^n|^{\frac{1}{n}}=\frac{5}{6}=L_1
  • \lim_{n\to \infty}|(\frac{1}{3})^n|^{\frac{1}{n}}=\frac{1}{3}=L_2

Therefore,

\lim_{n\to \infty}\frac{5^n+2^n}{6^n}=L_1+L_2=1.16>1

Hence by Cauchy root test given series is divergent.

5 0
4 years ago
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